Beamsplitter in a Michelson interferometer

[Julian Phillips]

For a science fair experiment at high school I built a Michelson interferometer that could achieve a single interference state.
Now there are two beams exit from such an interferometer, if a dielectric beamsplitter/combiner is used, one beam exits in a constructive state, and the other in a destructive state. If a metallic beamsplitter/combiner is used, they can be in the same state- so it was possible to null out the light in both beams. In this case the energy is absorbed at the beamsplitter/combiner and (I theorise) raises it's temperature (imagine what is happening to the electric currents the EM wave is inducing in the metallic coating) . It was a bit hard to measure any temperature rise with only 5mW of laser power.
So why could my interferometer with a metallic beamsplitter/combiner not create energy if tuned to only a constructive state ? Metallic beamsplitters are never 50/50. They always absorb enough energy to prevent that case happening- typically a little under 30%. Why this figure ? because if 1√ 2 is transmitted (70.7%) the waves will constructively interfere to produce 100%. So the energy that would normally be absorbed in the beamsplitter/combiner exits with less loss in the beamsplitter/combiner.

 

[sophiecentaur]

In this case the energy is absorbed at the beamsplitter/combiner

Do you mean that no light came out, direct or reflected? That sounds strange.
How do you "tune" a metallic beamsplitter? Do you have a vacuum coating facility to make your own?

 

[Julian Phillips]

Not quite 'no light' as the beamsplitter/combiner was not a perfect even split, but the light was greatly attenuated.

It is the interferometer that is tuned by adjusting the mirrors to align the beams and adjust the path length.
Once that was done the interferometer would slowly drift due to thermal effects causing the light to fade in and out over a period of tens of seconds.

The first metallic beamsplitter was made using chemical silver deposition (in 1993 and at school, that was the only means possible). Later on I was able to buy an inconel coated beamsplitter and repeat the experiment with better results.

This thread is reviving my interest in the experiment, and perhaps I should take it further with better equipment and a more powerful laser, and see if the beamsplitter temperature actually changes in relation to the interference state.

 

[Julian Phillips]

What is strange to me is that the means by which conservation of energy is achieved appears to be altered depending on the type of beamsplitter used.

 

[Charles Link]

@Julian Phillips It appears you are confusing the Fresnel coefficients (electric field amplitude ratios) with the energy reflection and transmission coefficient. (The energy is proportional to the second power of the E-field amplitude: Intensity $I=nE^2$ in simplified units). $\\$The concept for the case of an interferometer is slightly tricky. Suggestion is to read the Insights article https://www.physicsforums.com/insights/fabry-perot-michelson-interferometry-fundamental-approach/ In this article, it is explained how the energy transmission and reflection coefficients (surprisingly) can no longer be used for two beams, even though nothing changes physically in the beamsplitter when a second beam is introduced. It started out as a 50-50 (energy split) beamsplitter, but linear principles including superposition are not obeyed for the energy in this case, and don't need to be. $\\$ See also a thread where the metallic beamsplitter was discussed: https://www.physicsforums.com/threa...where-does-the-energy-go.942715/#post-5963655 $\\$ In both of these cases, we are in general treating the beamsplitters as having complete conservation of energy. The non-ideality of the metallic beamsplitter was also mentioned in post 26 of this thread. $\\$ The mathematics of the interferometer and the interference that occurs, although not readily obvious to most, is really quite simple, and also quite fascinating. Again, see the Insights article in the "link" above for a complete description.

 

[mfb]

I moved these posts into a separate thread as they distracted from the original question in the original thread.

 

[Julian Phillips]

@Charles Link Thanks Charles, the MIT video in the other thread was great to watch, and it is exactly what I observed when using a dielectric beamsplitter- and I came to the conclusion that it was the phase shifts at the beamsplitter interfaces causing that.
When using a metallic beamsplitter, both beams could be nulled- in this case do you think the energy appears as heat within the metallic coating ? It's not clear to me what the conclusion was in the other thread.

 

[Charles Link]

See https://arxiv.org/pdf/1509.00393.pdf Section 2.1 including the diagram there. Then also see post 5 and post 8 of https://www.physicsforums.com/threa...where-does-the-energy-go.942715/#post-5963655 There is some debate whether the $\frac{\pi}{2}$ phase shift in the metallic beamsplitter occurs on the transmitted waves or the reflected waves, but it needs to be introduced somewhere. The energy does not get nulled in both exit ports with this phase shift. When one exit port has the beams $\pi$ out of phase, they are in phase on the other exit port. $\\$ Having a $\frac{\pi}{2}$ relative phase between the two incident beams will cause this condition to occur. Reversing which one is ahead of the other will cause the entire output signal to shift from one exit port to the other. $\\$ The matrix formulation is just a simple way of writing the results for the two emerging beams using the Fresnel coefficients. Let me see if I can find a simple "link" that defines how the matrix works for this case...Yes. See section 2.3.1 of https://www.sciencedirect.com/topics/engineering/beam-splitter It's actually very simple. $\\$ Note that a $\frac{\pi}{2}$ phase shift is a factor of $e^{+i \pi/2}=i$, as you find for two of the 4 matrix elements shown in the paper of the first "link" above=Section 2.1.

 

[Julian Phillips]

..I'd better do the experiment again. To be honest, I think I'll be happier if I can't reproduce the original results.

 

[Charles Link]

..I'd better do the experiment again. To be honest, I think I'll be happier if I can't reproduce the original results.

The mirrors of the Michelson interferometer require careful alignment to get optimal results. Using corner cubes in place of mirrors can help to ease the alignment requirements.